need some help solving this problem:
$z^4*\overline{z}=-1+\sqrt{3}i$$\DeclareMathOperator{\cis}{cis}$
The answer is: $z_k=(2*{2^{1/5}})^{1/5}\cis(\frac{2\pi+6\pi*k}{15})$, $k=0,1,2,3,4$.
I tried using De Moivre to solve it but I got the wrong answer :
I substituted $z=r*\cis(\alpha)$
$r^4*\cis(4\alpha)*r*\cis(-\alpha)=2\cis(\frac{2\pi}{3})$
$r^5*\cis(3\alpha)=2\cis(\frac{2\pi}{3})$
$r*\cis(\frac{3\alpha}{5})=2^{1/5}\cis(\frac{\frac{2\pi}{3}+2\pi*k}{5})=2^{1/5}\cis(\frac{2\pi+6\pi*k}{15})$
from here I just couldn't find a way to get to the right answer.
Since $r^5\operatorname{cis}3\alpha=2\operatorname{cis}\frac{2\pi(1+3n)}{3}$ with $n\in\Bbb Z$, $r=2^{1/5}$ while $\alpha=\frac{2\pi(1+3n)}{9}$ with $0\le n\le4$.